A step in a proof that a set is Lebesgue measurable if and only if its translation is

Question

I'm trying to fill in the gaps in the following proof:

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In the above, $\mu^*$ refers to the Lebesgue outer measure, the Carathéodory definition of measurability states that a set $A$ is Carathéodory measurable if for any subset $E$ of $\mathbb{R}^n$ one has $$\mu^*(E)=\mu^*(E\cap A) + \mu^*(E\cap A^c),$$ and a set is defined as measurable if it belongs to the $\sigma$-algebra formed by all the Carathédory measurable subsets of $\mathbb{R}^n$ (equivalently, if it is Carathéodory measurable).

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I understand why $\mu^*(A+h)=\mu^*(A)$, but I fail to understand why a set $A$ is measurable if and only if so is $A+h$. What we wish to prove is that, if $A$ is measurable, then for any subset $E$ of $\mathbb{R}^n$ we have that $$\mu^*(E)=\mu^*(E\cap (A+h)) + \mu^*(E\cap (A+h)^c),$$ but I do not see how the fact that $$(E+h)\cap (A+h) = (E\cap A) + h$$ is helpful here.

Answer 1

Let $A, E \subset \mathbb{R}^n$ be arbitrary. Using $(A + h)^c = A^c + h$, we have \begin{align} \mu^*((E + h) \cap (A + h)) + \mu^*((E + h) \cap (A + h)^c) &= \mu^*(E \cap A + h) + \mu^*(E \cap A^c + h) \\ &= \mu^*(E \cap A) + \mu^*(E \cap A^c). \end{align}

Now for your problem, to check measurability of $A + h$, you can check whether $\mu^*(E + h) = \mu^*((E + h) \cap (A + h)) + \mu^*((E + h) \cap (A + h)^c)$ for all $E$ since $E \mapsto E + h$ is a bijection.

Answer 2

It's a fiddly calculation with sets. Sketch: for convenience, drop the star and replace $\mu$ by $m.$ Also, following standard notation, take the measurable set to be $E$ and consider an arbitrary set $A.$ Then

$x \in A\cap (E+h),\Leftrightarrow x-h\in (A-h)\cap E\Leftrightarrow x\in (A-h)\cap E+h\Rightarrow$

$\tag1 A\cap (E+h)=(A-h)\cap E+h$

Similarly,

$\tag2 A\cap (E + h)^c=(A − h)\cap E^c+h.$

Now use the translation invariance of $m$ together with the definition of measurability.